[tex]\mathbf a=(1,1,1)^\top=1\,\mathbf i+1\,\mathbf j+1\,\mathbf k[/tex]
[tex]\mathbf b=(-2,3,0)^\top=-2\,\mathbf i+3\,\mathbf j[/tex]
[tex]\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\-2&3&0\end{vmatrix}=-3\,\mathbf i-2\,\mathbf j+5\,\mathbf k=(-3,-2,5)^\top[/tex]
Basically, you're looking for a matrix [tex]\mathbf A[/tex] such that
[tex]\mathbf A(\mathbf a\times\mathbf b)=\mathbf 0[/tex]
i.e. a matrix [tex]\mathbf A[/tex] whose nullspace with basis vector [tex]\mathbf a\times\mathbf b[/tex].
By the rank-nullity theorem, the rank of [tex]\mathbf A[/tex] and the dimension of its nullspace must add up to the number of columns, so
[tex]\mathrm{rank}\mathbf A+\underbrace{\mathrm{null}\mathbf A}_1=3\implies\mathrm{rank}\mathbf A=2[/tex]
One easy choice for a row would be [tex]\begin{bmatrix}1&1&1\end{bmatrix}[/tex], since
[tex](1,1,1)(-3,-2,5)^\top=0[/tex]
Now you only need to find another combination such that the second row of [tex]\mathbf A[/tex] is independent of the first. An easy choice for this is to let the first element be 0, and the next be 1. Then the last element must be [tex]\dfrac25[/tex], as
[tex]\left(0,1,\dfrac25\right)(-3,-2,5)^\top=0[/tex]
So,
[tex]\underbrace{\begin{bmatrix}1&1&1\\0&1&\frac25\end{bmatrix}}_{\mathbf A}\begin{bmatrix}-3\\-2\\5\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}[/tex]
is one possible solution.
